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Slide 1 Chapter 3 Particle-in-Box (PIB) Models. Slide 2 Carrots are Orange. Tomatoes are Red. But Why?

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Presentation on theme: "Slide 1 Chapter 3 Particle-in-Box (PIB) Models. Slide 2 Carrots are Orange. Tomatoes are Red. But Why?"— Presentation transcript:

1 Slide 1 Chapter 3 Particle-in-Box (PIB) Models

2 Slide 2 Carrots are Orange. Tomatoes are Red. But Why?

3 Slide 3 Outline The Classical Particle in a Box (PIB) The Quantum Mechanical PIB PIB and the Color of Vegetables PIB Properties Multidimensional Systems: The 2D PIB and 3D PIB Other PIB Models The Free Electron Molecular Orbital (FEMO) Model Statistical Thermodynamics: Translational contributions to the thermodynamic properties of gases.

4 Slide 4 The Classical Particle in a Box x P(x) 0 a 0 a E = ½mv 2 = p 2 /2m = any value, including 0 i.e. the energy is not quantized and there is no minimum. P(x) = Const. = C 0 x a P(x) = 0 x, x > a

5 Slide 5 P(x) = Const. = C 0 x a P(x) = 0 x, x > a Normalization Note:

6 Slide 6 Outline The Classical Particle in a Box (PIB) The Quantum Mechanical PIB PIB and the Color of Vegetables PIB Properties Multidimensional Systems: The 2D PIB and 3D PIB Other PIB Models The Free Electron Molecular Orbital (FEMO) Model Statistical Thermodynamics: Translational contributions to the thermodynamic properties of gases.

7 Slide 7 Quantum Mechanical Particle-in-Box x V(x) 0 a 0 V(x) = 0 0 x a V(x) x a Outside the box: x a P(x) = *(x) (x) = 0 (x) = 0 Inside the box: 0 x a Schrödinger Equation or

8 Slide 8 Solving the Equation Assume So far, there is no restriction on and, hence none on the energy, E (other than that it cannot be negative)

9 Slide 9 Applying Boundary Conditions (BCs) Nature hates discontinuous behavior; i.e. Its not really possible to stop on a dime Because the wavefunction,, is 0 outside the box, x a, it must also be 0 inside the box at x=0 and x=a BC-1: (0) = 0 = Asin(0) + Bcos(0) = B Therefore, (x) = Asin( x) BC-2: (a) = 0 = Asin( a) sin(n ) = 0 Therefore: a = n n = 1, 2, 3,... Why isnt n = 0 acceptable? or:n = 1, 2, 3,...

10 Slide 10 Wavefunctions and Energy Levels Note that, unlike the classical particle in a box: (a) the allowed energies are quantized (b) E = 0 is NOT permissible (i.e. the particle cant stand still) n = 1, 2, 3,...

11 Slide 11 Energy Quantization results from application of the Boundary Conditions

12 Slide 12 0a 2 0a Wavefunctions

13 Slide 13 Note that: (a) The Probability, 2 is very different from the classical result. (b) The number of nodes increases with increasing quantum number. The increasing number of nodes reflects the higher kinetic energy with higher quantum number.

14 Slide 14 The Correspondence Principle The predictions of Quantum Mechanics cannot violate the results of classical mechanics on macroscopically sized systems. Consider an electron in a 1 Angstrom box. Calculate (a) the Zero Point Energy (i.e. minimum energy) (b) the minimum speed of the electron m e = 9.1x kg a = 1x m h = 6.63x J-s Thus, the minimum speed of an electron confined to an atom is quite high.

15 Slide 15 Consider a 1 gram particle in a 10 cm box. Calculate (a) the Zero Point Energy (i.e. minimum energy) (b) the minimum speed of the particle m = 1x10 -3 kg a = 0.10 m h = 6.63x J-s Thus, the minimum energy and speed of a macroscopic particle are completely negligible. Let's perform the same calculation on a macroscopic system.

16 Slide 16 Probability Distribution of a Macroscopic Particle Consider a 1 gram particle in a 10 cm box moving at 1 cm/s. Calculate the quantum number, n, which represents the number of maxima in the probability, 2. m = 1x10 -3 kg a = 0.10 m v = 0.01 m/s h = 6.63x J-s Thus, the probability distribution is uniform throughout the box, as predicted by classical mechanics.

17 Slide 17 Outline The Classical Particle in a Box (PIB) The Quantum Mechanical PIB PIB and the Color of Vegetables PIB Properties Multidimensional Systems: The 2D PIB and 3D PIB Other PIB Models The Free Electron Molecular Orbital (FEMO) Model Statistical Thermodynamics: Translational contributions to the thermodynamic properties of gases.

18 Slide 18 Some Useful Integrals

19 Slide 19 PIB Properties Normalization of the Wavefunctions 0 x a

20 Slide 20 Probability of finding the particle in a particular portion of the box This is significantly lower than the classical probability, Calculate the probability of finding a particle with n=1 in the region of the box between 0 and a/4

21 Slide 21 Orthogonality of the Wavefunctions Thus, 1 and 2 are orthogonal Two wavefunctions are orthogonal if: We will show that the two lowest wavefunctions of the PIB are orthogonal:

22 Slide 22 Thus, the PIB wavefunctions are orthogonal. If they have also been normalized, then the wavefunctions are orthonormal: Using the same method as above, it can be shown that: or

23 Slide 23 Positional Averages We will calculate the averages for 1 and present the general result for n

24 Slide 24 This makes sense because 2 is symmetric about the center of the box. General Case:

25 Slide 25

26 Slide 26 Note that the general QM value reduces to the classical result in the limit of large n, as required by the Correspondence Principle. compared to the classical result General Case:

27 Slide 27 Momentum Averages Preliminary ^

28 Slide 28 = 0 On reflection, this is not surprising. The particle has equal probabilities of moving in the + or - x-direction, and the momenta cancel each other. (General case is the same)

29 Slide 29 ^ ^ General Case:

30 Slide 30 Standard Deviations and the Uncertainty Principle Heisenberg Uncertainty Principle:

31 Slide 31 Outline The Classical Particle in a Box (PIB) The Quantum Mechanical PIB PIB and the Color of Vegetables PIB Properties Multidimensional Systems: The 2D PIB and 3D PIB Other PIB Models The Free Electron Molecular Orbital (FEMO) Model Statistical Thermodynamics: Translational contributions to the thermodynamic properties of gases.

32 Slide 32 Other PIB Models PIB with one non-infinite wall x V(x) 0 a 0 V(x) = 0 0 x a V(x) x < 0 VoVo V(x) = V o x > a We will first consider the case where E < V 0. Region I I Region II II Boundary Conditions The wavefunction must satisfy two conditions at the boundaries (x=0, x=a, x ) 1) must be continuous at all boundaries. 2) d /dx must be continuous at the boundaries unless V at the boundary.

33 Slide 33 x V(x) 0 a 0 V(x) = 0 0 x a V(x) x < 0 VoVo V(x) = V o x > a We will first consider the case where E < V 0. Region I I Region II II Region I can assume

34 Slide 34 x V(x) 0 a 0 V(x) = 0 0 x a V(x) x < 0 VoVo V(x) = V o x > a We will first consider the case where E < V 0. Region I I Region II II Region I Relation between and E

35 Slide 35 x V(x) 0 a 0 V(x) = 0 0 x a V(x) x < 0 VoVo V(x) = V o x > a We will first consider the case where E < V 0. Region I I Region II II Region II can assume

36 Slide 36 x V(x) 0 a 0 V(x) = 0 0 x a V(x) x < 0 VoVo V(x) = V o x > a We will first consider the case where E < V 0. Region I I Region II II Region II Relation between and E

37 Slide 37 x V(x) 0 a 0 V(x) = 0 0 x a V(x) x < 0 VoVo V(x) = V o x > a We will first consider the case where E < V 0. Region I I Region II II BC: x = 0 Boundary Conditions Therefore:

38 Slide 38 x V(x) 0 a 0 V(x) = 0 0 x a V(x) x < 0 VoVo V(x) = V o x > a We will first consider the case where E < V 0. Region I I Region II II BC: x = Boundary Conditions Therefore:

39 Slide 39 x V(x) 0 a 0 V(x) = 0 0 x a V(x) x < 0 VoVo V(x) = V o x > a We will first consider the case where E < V 0. Region I I Region II II BCs: x = a Boundary Conditions or

40 Slide 40 x V(x) 0 a 0 V(x) = 0 0 x a V(x) x < 0 VoVo V(x) = V o x > a We will first consider the case where E < V 0. Region I I Region II II These equations can be solved analytically to obtain the allowed values of the energy, E. However, its fairly messy. Ill just show you some results obtained by numerical solution of the Schrödinger Equation. BCs: x = a Boundary Conditions

41 Slide 41 x 2 x x 2 x x 2 x This is an arbitrary value of V 0

42 Slide 42 x x 2 What happened?? Why is the wavefunction so different? The condition for the earlier solution, E < V 0, is no longer valid.

43 Slide 43 0 a 0 VoVo Region I I Region II II V(x) = 0 0 x a V(x) x < 0 V(x) = V o x > a Region I Region II E > V o

44 Slide 44 PIB with central barrier: Tunneling Preliminary: Potential Energy Barriers in Classical Mechanics V 0 = mgh = 1430 J Bowling Ball m = 16 lb = 7.3 kg v = 30 mph = 13.4 m/s h = 20 m KE = ½mv 2 = 650 J Will the bowling ball make it over the hill?Of course not!! In classical mechanics, a particle cannot get to a position in which the potential energy is greater than the particles total energy.

45 Slide 45 x V(x) 0 0 x1x1 VoVo x2x2 a I II III V(x) x < 0 V(x) = 0 0 x x 1 V(x) x > a V(x) = V 0 x 1 x x 2 V(x) = 0 x 2 x a Ill just show the Boundary Conditions and graphs of the results. BC: x=0BCs: x=x 1 BCs: x=x 2 BC: x=a

46 Slide 46 x 2 x Note that there is a significant probability of finding the particle inside the barrier, even though V 0 > E. This means that the particle can get from one side of the barrier to the other by tunneling through the barrier.

47 Slide 47 We have just demonstrated the quantum mechanical phenomenon called tunneling, in which a particle can be in a region of space where the potential energy is higher than the total energy of the particle. This is not simply an abstract phenomenon, but is known to occur in many areas of Chemistry and Physics, including: Ammonia inversion (the ammonia clock) Kinetic rate constants Charge carriers in semiconductor devices Nuclear radioactive decay Scanning Tunneling Microscopy (STM)

48 Slide 48 Outline The Classical Particle in a Box (PIB) The Quantum Mechanical PIB PIB and the Color of Vegetables PIB Properties Multidimensional Systems: The 2D PIB and 3D PIB Other PIB Models The Free Electron Molecular Orbital (FEMO) Model Statistical Thermodynamics: Translational contributions to the thermodynamic properties of gases.

49 Slide 49 Carrots are Orange. Tomatoes are Red. But why??

50 Slide 50 Free Electron Molecular Orbital (FEMO) Model So, does the solution to the quantum mechanical particle in a box serve any role other than to torture P. Chem. students??? Yes!! To a good approximation, the electrons in conjugated polyalkenes are free to move within the confines of the orbital system. a Notes: Estimation of the box length, a, will be discussed later. Each carbon in this conjugated system contributes 1 electron. Thus, there are 6 electrons in the -system of 1,3,5-hexatriene (above)

51 Slide 51 Bonding in Ethylene Bonding ( ) Orbital Anti-bonding ( *) Orbital Maximum electron density between Cs PIB ( 1 2 ) Maximum electron density between Cs Electron density node between Cs PIB ( 2 2 ) Electron density node between Cs

52 Slide 52 * Transition in Ethylene using FEMO h = 6.63x Js m = 9.11x kg a = ??? R = nm R ½R½R ½R½R a = 2R It is common (although not universal) to assume that the electrons are free to move approximately ½ bond length beyond each outermost carbon. = nm

53 Slide 53 Calculation of max = 7.9x10 -8 m 80 nm (exp) = 190 nm The difficulty with applying FEMO to ethylene is that the result is extremely sensitive to the assumed box length, a. Units:

54 Slide 54 Application of FEMO to 1,3-Butadiene R ½R½R ½R½RRR It is common to use R = 0.14 nm (the C-C bond length in benzene,where the bond order is 1.5) as the average bond length in conjugated polyenes. L = 3R + 2(½R) = 4R L = nm = 0.56 nm

55 Slide 55 Calculation of max = 2.07x10 -7 m 207 nm (exp) = 217 nm

56 Slide 56 Application of FEMO to 1,3,5-Hexatriene a = 5R + 2(½R) = 6R a = nm = 0.84 nm R ½R½R ½R½R RR RR E = 5.98x J = 332 nm (exp) = 258 nm Problem worked out in detail in Chapter 3 HW

57 Slide 57 The Box Length a = 5R + 2(½R) = 6R R ½R½R ½R½R RR RR Hexatriene General: a = nbR + 2(½R) =(nb+1)RR = 1.40 Å = 0.14 nm

58 Slide 58 Outline The Classical Particle in a Box (PIB) The Quantum Mechanical PIB PIB and the Color of Vegetables PIB Properties Multidimensional Systems: The 2D PIB and 3D PIB Other PIB Models The Free Electron Molecular Orbital (FEMO) Model Statistical Thermodynamics: Translational contributions to the thermodynamic properties of gases.

59 Slide 59 PIB and the Color of Vegetables Introduction The FEMO model predicts that the * absorption wavelength,, increases with the number of double bonds, #DB Compound #DB (FEMO) Ethylene1 80 nm Butadiene2207 Hexatriene3333 roughly N = # of electron pairs This is because: and:

60 Slide 60 Blue Yellow Red (nm) % Transmission Blue Yellow Red (nm) % Transmission If a substance absorbs light in the blue region of the visible spectrum, the color of the transmitted (or reflected) light will be red. If a substance absorbs light in the red region of the visible spectrum, the color of the transmitted (or reflected) light will be blue. Light Absorption and Color

61 Slide 61 Conjugated Systems and the Color of Substances Ethylene nm UV 1,3-Butadiene UV 1,3,5-Hexatriene UV -Carotene Vis. Lycopene Vis. White light contains all wavelengths of visible radiation ( = nm) If a substance absorbs certain wavelengths, then the remaining light is reflected, giving the appearance of the complementary color. e.g. a substance absorbing violet light appears to be yellow in color. Carrots Tomatoes No. of Conj. Molecule Doub. Bonds max Region

62 Slide 62 Structure of Ground State Butadiene R(C-C) = 1.53 Å 1.32 Å 1.47 Å Calculation: HF/6-31G(d) – 1 minute As well known, there is a degree of electron delocalization between the conjugated double bonds, giving some double bond character to the central bond.

63 Slide 63 Calculation: HF/6-31G(d) – 37 optimizations – 25 minutes Potential Energy Surface (PES) of Ground State Butadiene Minimum at ~40 o

64 Slide 64 Energy Difference Between Trans and Cis-Butadiene E(cal) = E cis – E trans = 12.7 kJ/mol Experiment At room temperature (298 K), butadiene is a mixture with approximately 97% of the molecules in the trans conformation (and 3% cis). We could get very close to experiment by using a higher level of theory.

65 Slide 65 Structure of Excited State Butadiene 1.32 Å 1.47 Å Ground State: S Å 1.39 Å Excited State: S 1

66 Slide 66 Structure of Excited State Butadiene 1.32 Å 1.47 Å Ground State: S Å 1.39 Å Excited State: S 1 HOMO LUMO * Transition

67 Slide 67 Comparison of FEMO and QM Wavefunctions 0L HOMO LUMO

68 Slide 68 Outline The Classical Particle in a Box (PIB) The Quantum Mechanical PIB PIB and the Color of Vegetables PIB Properties Multidimensional Systems: The 2D PIB and 3D PIB Other PIB Models The Free Electron Molecular Orbital (FEMO) Model Statistical Thermodynamics: Translational contributions to the thermodynamic properties of gases.

69 Slide 69 Multidimensional Systems: The 2D PIB The Potential Energy 0a 0 b x y V(x,y) = 0 0 x a and 0 y b V(x,y) x a or y b Outside of the box i.e. x a or y b (x,y) = 0 The 2D Hamiltonian: The 2D Schr. Eqn.:

70 Slide 70 Solution: Separation of Variables Inside the box: V(x,y) = 0 0 x a and 0 y b The 2D Schr. Eqn.: Note: On following slides, I will show how this two dimensional differential equation can be solved by the method of Separation of Variables. You are NOT responsible for the details of the soution, but only the assumptions and results.

71 Slide 71 Solution: Separation of Variables HxHx HyHy Inside the box: Assume:

72 Slide 72 Solution: Separation of Variables If: f(x) + g(y) = C then: f(x) = C 1 and g(y) = C 2 C 1 + C 2 = C f(x)g(y)C = ExEx = EyEy E = E x + E y Inside the box:

73 Slide 73 = ExEx = EyEy Range 0 x a Range 0 y b E = E x + E y

74 Slide 74 Two Dimensional PIB: Summary Inside the box: Assume: Range 0 x a Range 0 y b E = E x + E y

75 Slide 75 n x = 1, 2, 3,... n y = 1, 2, 3,... Range 0 x a 0 y b n x = 1 n y = 1 2 n x = 2 n y = 1 2 n x = 2 n y = 2 2 The Wavefunctions

76 Slide 76 The Energies: Wavefunction Degeneracy n x = 1, 2, 3,... n y = 1, 2, 3,... Square Box a = b n x n y g = g = g = g = 2

77 Slide 77 Application: * Absorption in Benzene 1 n x n y E [h 2 /8ma 2 ] The six electrons in benzene can be approximated as particles in a square box. One can estimate the wavelength of the lowest energy * from the 2D-PIB model (See HW problem) n x n y E [h 2 /8ma 2 ]

78 Slide 78 a b c V(x,y,z) = 0 0 x a 0 y b 0 z c V(x,y,z) Outside the box Three Dimensional PIB

79 Slide 79 Assume:

80 Slide 80 n x = 1, 2, 3,... n y = 1, 2, 3,... n z = 1, 2, 3,... Cubical Box a = b = c Note: You should be able to determine the energy levels and degeneracies for a cubical box and for various ratios of a:b:c

81 Slide 81 Outline The Classical Particle in a Box (PIB) The Quantum Mechanical PIB PIB and the Color of Vegetables PIB Properties Multidimensional Systems: The 2D PIB and 3D PIB Other PIB Models The Free Electron Molecular Orbital (FEMO) Model Statistical Thermodynamics: Translational contributions to the thermodynamic properties of gases.

82 Slide 82 Introduction to Statistical Thermodynamics But this is a Quantum Mechanics course. Why discuss thermodynamics?? Statistical Thermodynamics is the application of Statistical Mechanics to predict thermodynamic properties of molecules. 1. The Statistical Thermodynamic formulas used to predict the properties come from the Quantum Mechanical energies for systems such as the PIB, Rigid Rotor (Chap. 4) and Harmonic Oscillator (Chap. 5). 2. Various properties of the molecules, including moments of inertia, vibrational frequencies, and electronic energy levels are required to calculate the thermodynamic properties Often, these properties are not available experimentally, and must be obtained from QM calculations.

83 Slide 83 Some Applications of QM/Stat. Thermo. Molecular Heat Capacities [C V (T) and C P (T)] Molecular Enthalpies of Formation [ H f ] Reaction Enthalpy Changes [ H rxn ] Reaction Entropy Changes [ S rxn ] Reaction Gibbs Energy Changes [ G rxn ] and Equilibrium Constants [K eq ] Kinetics: Activation Enthalpies [ H ] and Entropies [ S ] Kinetics: Rate Constants Bond Dissociation Enthalpies [aka Bond Strengths]

84 Slide 84 Frequencies Thermochemistry Temperature Kelvin. Pressure Atm. Thermochemistry will use frequencies scaled by Atom 1 has atomic number 8 and mass Atom 2 has atomic number 8 and mass Molecular mass: amu. Principal axes and moments of inertia in atomic units: EIGENVALUES X Y Z THIS MOLECULE IS A PROLATE SYMMETRIC TOP. ROTATIONAL SYMMETRY NUMBER 2. ROTATIONAL TEMPERATURE (KELVIN) ROTATIONAL CONSTANT (GHZ) Zero-point vibrational energy (Joules/Mol) (Kcal/Mol) VIBRATIONAL TEMPERATURES: (KELVIN) Zero-point correction= (Hartree/Particle) Thermal correction to Energy= Thermal correction to Enthalpy= Thermal correction to Gibbs Free Energy= Sum of electronic and zero-point Energies= Sum of electronic and thermal Energies= Sum of electronic and thermal Enthalpies= Sum of electronic and thermal Free Energies= E (Thermal) CV S KCAL/MOL CAL/MOL-K CAL/MOL-K TOTAL ELECTRONIC TRANSLATIONAL ROTATIONAL VIBRATIONAL Q LOG10(Q) LN(Q) TOTAL BOT D TOTAL V= D VIB (BOT) D VIB (V=0) D ELECTRONIC D TRANSLATIONAL D ROTATIONAL D Output from G-98 geom. opt. and frequency calculation on O 2 (at 298 K) QCISD/6-311G(d)

85 Slide 85 Mini-Outline for Stat. Thermo. Development The Boltzmann Distribution The Partition Function (Q) Thermodynamic properties in terms of Q Partition function for non-interacting molecules The molecular partition function (q) The total partition function for a system of molecules (Q) The translational partition function Translational contributions to thermodynamic properties

86 Slide 86 The Boltzmann Distribution E0E0 E1E1 E2E2 Normalization Therefore: Boltzmann Distribution

87 Slide 87 The Partition Function Q is called the "Partition Function". Thermodynamic properties, including, U, C v, H, S, A, G, can be calculated from Q. Alternative Form One can sum over energy "levels" rather than individual energy states. g j is the degeneracy of the jth energy level, E j where

88 Slide 88 Thermodynamic Properties from Q: Internal Energy (U) Derivative at constant V,N because energy depends on volume and No. of Molecules NOTE: On an exam, you will be provided with any required SM formula for a thermodynamic property (e.g. the above expression for U)

89 Slide 89 Thermodynamic Properties from Q: Other Properties The math involved in relating other thermodynamic properties to Q is a bit more involved. It can be found in standard P. Chem. texts; e.g. Physical Chemistry, by R. A. Alberty and R. J. Silbey, Chap. 17 I'll just give the results. Constant Volume Heat Capacity: Pressure: Entropy:

90 Slide 90 Thermodynamic Properties from Q: Other Properties Note: If Q f(V), then H = U Note: If Q f(V), then G = A Enthalpy: Constant Pressure Heat Capacity: Helmholtz Energy: Gibbs Energy: NOTE: On an exam, you will be provided with any required SM formula for a thermodynamic property.

91 Slide 91 Partition Function for Non-interacting Molecules (aka IDG) Case A: Distinguishable Molecules The microstate of molecule 'a' is given by j, of molecule 'b' by k… If the molecules are the same, Therefore, for N molecules we have: a, b, c,... designates the molecule i, j, k,... designates the molecule's set of energies (e.g. trans., rot., vib.,...)

92 Slide 92 Partition Function for Non-interacting Molecules (aka IDG) Case B: Indistinguishable Molecules In actuality, one cannot distinguish between different molecules in a gas. Let's say that there is one state in which the quantum numbers for molecule 'a' are m 1,n 1 … and for molecule 'b' are m 2,n 2 … This state is the same as one in which the quantum numbers for molecule 'a' are m 2,n 2 … and for molecule 'b' are m 1,n 1 … However, the two sets of quantum numbers above would give two terms in the above equation for Q. i.e. we've overcounted the number of terms. It can be shown that the overcounting can be eliminated by dividing Q by N! (the number of ways of permuting N particles). Therefore, Indistinguishable Molecules: Distinguishable Molecules:

93 Slide 93 The Molecular Partition Function (q) To a good approximation, the Hamiltonian for a molecule can be written as the sum of translational, rotational, vibrational and electronic Hamiltonians: Therefore, the molecule's total energy ( i ) is the sum of individual energies: The partition function is:

94 Slide 94 The Total Partition Function (Q) It can be shown that the term, 1/N! belongs to Q tran

95 Slide 95 Thermodynamic Properties of Molecules The expressions for the thermodynamic properties all involve lnQ; e.g. where Similarly etc.

96 Slide 96 The Translational Partition Function for an Ideal Gas Consider that the molecules of a gas are confined in a cubical box of length 'a' on each side.* *One gets the same result if the box is not assumed to be cubical, with a bit more algebra. The energy levels are given by: The molecular translational partition function is:

97 Slide 97 One can show that the summand (term in summation) is an extremely slowly varying function of n. For example, for O 2 [M = 32x10 -3 kg/mol] in a 10 cm (0.1 m) box at 298 K: n Δ[Expon] 1 8x x x10 -8 This leads to an important simplification in the expression for q tran. The extremely slowly changing exponent results from the fact that: is the change in energy between successive quantum levels.

98 Slide nini f(n i ) Relation between sum and integral for slowly varying functions If f(n i ) is a slowly varying function of n i, then one may replace the sum by the integral:

99 Slide 99 V is the volume of the box Note: The translation partition function is the only one which depends upon the systems volume.

100 Slide 100 This is the translational partitition function for a single molecule. A note on the calculation of V (in SI Units) One is normally given the temperature and pressure, from which one can calculate V from the IDG law. One should use the pressure in Pa [1 atm. = 1.013x10 5 Pa], in which case V is given in m 3 [SI Units] R = 8.31 J/mol-K = 8.31 Pa-m 3 /mol-K

101 Slide 101 Calculate q tran for one mole of O 2 at 298 K and 1 atm. N A = 6.02x10 23 mol -1 h = 6.63x J-s k = 1.38x J/K R = 8.31 Pa-m 3 /mol-K M = 32 g/mol 1 J = 1 kg-m 2 /s 2 1 atm. = 1.013x10 5 Pa

102 Slide Thermochemistry Temperature Kelvin. Pressure Atm. Q LOG10(Q) LN(Q) TOTAL BOT D TOTAL V= D VIB (BOT) D VIB (V=0) D ELECTRONIC D TRANSLATIONAL D ROTATIONAL D G-98 Output Units: Their output is actually q tran / N A

103 Slide 103 Partition Function for a system with N (=nN A ) molecules This is the translational partitition function for a single molecule. Stirlings Approximation

104 Slide 104 This is an extremely large number. However, as well see shortly, ln(Q) is multiplied by the very small number (k) in the thermodynamics formulas. Calculate ln(Q tran ) for one mole of O 2 at 298 K and 1 atm. q tran = 4.28x10 30

105 Slide 105 Translational Contributions to the Thermodynamic Properties of Ideal Gases Preliminary

106 Slide 106 The Ideal Gas Equation Notes: (1) Because none of the other partition functions (rotational, vibrational, or electronic) depend on V, they do not contribute to the pressure. (2)We end up with the IDG equation because we assumed that the molecules dont interact when we assumed that the total energy is the sum of individual molecules energies. (because R = kN A )

107 Slide 107 Internal Energy This demonstrates the Principle of Equipartition of Translational Internal Energy (1/2RT per translation) It arose from the assumption that the change in the exponent is very small and the sum can be replaced by an integral. This is always true for translational motions of gases. or Molar Internal Energy

108 Slide 108 Enthalpy Note: As noted earlier, if Q is independent of V, then H = U. This is the case for all partition functions except Q tran or Molar Enthalpy

109 Slide 109 Heat Capacities Experimental Heat Capacities at K Compd. C P (exp) Ar J/mol-K O SO

110 Slide 110 E (Thermal) CV S KCAL/MOL CAL/MOL-K CAL/MOL-K TOTAL ELECTRONIC TRANSLATIONAL ROTATIONAL VIBRATIONAL Q LOG10(Q) LN(Q) TOTAL BOT D TOTAL V= D VIB (BOT) D VIB (V=0) D ELECTRONIC D TRANSLATIONAL D ROTATIONAL D Output from G-98 geom. opt. and frequency calculation on O 2 (at 298 K) QCISD/6-311G(d) U tran (mol) = 3/2RT = 3.72 kJ/mol = kcal/mol C V tran (mol) = 3/2R = J/mol-K = 2.98 cal/mol-K

111 Slide 111 ln(Q tran ) = 1.01x10 25 mol -1 Earlier, we showed that for one mole of O 2 at 298 K and 1 atm., O 2 : S mol (exp) = J/mol-K at K Entropy

112 Slide 112 E (Thermal) CV S KCAL/MOL CAL/MOL-K CAL/MOL-K TOTAL ELECTRONIC TRANSLATIONAL ROTATIONAL VIBRATIONAL Q LOG10(Q) LN(Q) TOTAL BOT D TOTAL V= D VIB (BOT) D VIB (V=0) D ELECTRONIC D TRANSLATIONAL D ROTATIONAL D Output from G-98 geom. opt. and frequency calculation on O 2 (at 298 K) QCISD/6-311G(d) G-98: S mol tran = cal/mol-K = J/mol-K Difference is round-off error and less Sig. Figs. in our calculation. Our Calculation (above): S mol tran = J/mol-K

113 Slide 113 Helmholtz Energy ln(Q tran ) = 1.01x10 25 mol -1 For O 2 at 298 K (one mol): Gibbs Energy For O 2 at 298 K (one mol): Note: As noted before, if Q is independent of V, then G = A. This is the case for all partition functions except Q tran

114 Slide 114 Frequencies Thermochemistry Temperature Kelvin. Pressure Atm. Thermochemistry will use frequencies scaled by Atom 1 has atomic number 8 and mass Atom 2 has atomic number 8 and mass Molecular mass: amu. Principal axes and moments of inertia in atomic units: EIGENVALUES X Y Z THIS MOLECULE IS A PROLATE SYMMETRIC TOP. ROTATIONAL SYMMETRY NUMBER 2. ROTATIONAL TEMPERATURE (KELVIN) ROTATIONAL CONSTANT (GHZ) Zero-point vibrational energy (Joules/Mol) (Kcal/Mol) VIBRATIONAL TEMPERATURES: (KELVIN) Zero-point correction= (Hartree/Particle) Thermal correction to Energy= Thermal correction to Enthalpy= Thermal correction to Gibbs Free Energy= Sum of electronic and zero-point Energies= Sum of electronic and thermal Energies= Sum of electronic and thermal Enthalpies= Sum of electronic and thermal Free Energies= Output from G-98 geom. opt. and frequency calculation on O 2 (at 298 K) QCISD/6-311G(d) These are thermal contributions to U, H and G. They represent the sum of the translational, rotational, vibrational and (in some cases) electronic contributions. They are given in atomic units (au). 1 au = kJ/mol

115 Slide 115 Translational Contributions to O 2 Entropy Obviously, there are significant additional contributions (future chapters)

116 Slide 116 Translational Contributions to O 2 Enthalpy Obviously, there are significant additional contributions (future chapters)

117 Slide 117 Translational Contributions to O 2 Heat Capacity Obviously, there are significant additional contributions (future chapters)


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